Find $\dfrac{d}{dx}(-3\cdot2^x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $2^x\cdot (-3\log_2(x))$ (Choice B) B $2^x\cdot (-3\ln(2))$ (Choice C) C $-6^{x-1}$ (Choice D) D $-6^x\cdot(\ln(x))$
The expression to differentiate includes an exponential term. Remember that the derivative of the general exponential term $a^x$ (where $a$ is any positive constant) is $\ln(a)\cdot a^x$. Put another way, $\dfrac{d}{dx}(a^x)=\ln(a)\cdot a^x$. $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(-3\cdot2^x) \\\\ &=-3\dfrac{d}{dx}(2^x) \\\\ &=-3\cdot\ln(2)\cdot2^x \\\\ &=2^x\cdot(-3\ln(2)) \end{aligned}$ In conclusion, $\dfrac{d}{dx}(-3\cdot2^x)=2^x\cdot(-3\ln(2))$.